Folliation and non-vanishing vector field. The canonical foliation on $\mathbb{R}^k$ is its decomposition into parallel sheets $\{t\} \times \mathbb{R}^{k-1}$ (as oriented submanifolds). In general, a foliation $\mathcal{F}$ on a compact, oriented manifold $X$ is a decomposition into $1-1$ immersed oriented manifolds $Y_\alpha$ (not necessarily compact) that is locally given (preserving all orientations) by the canonical foliation in a suitable chart at each point. For example, the lines in $\mathbb{R}^2$ of any fixed slope (possibly irrational) descend to a foliation on $T^2 = \mathbb{R}^2/\mathbb{Z}^2$.
(a) If $X$ admits a foliation, prove that $\chi(X) = 0$. (Hint: Partition of unity.)
(b) Prove (with suitable justification) that $S^2 \times S^2$ does not admit a foliation as defined above.

Theorem
  A compact, connected, oriented manifold $X$ possesses a nowhere vanishing vector field if and only if its Euler characteristic is zero.

Question: How could $X$ in this problem satisfy the connectness property in the theorem? Can I just say if it is not connected, treat each connected component individually?
 A: I assume that your manifold and foliation are smooth and foliation is of codimension 1, otherwise see Jack Lee'a comment. Then pick a Riemannian metric on $X$ and at each point $x\in M$ take unit vector $u_x$ orthogonal to the leaf $F_x$ through $x$: There are two choices, but since your foliation is transversally orientable, you can make a consistent choice of $u_x$. Then $u$ is a nonvanishing vector field on $X$. 
In fact, orientability is irrelevant: Clearly, it suffices to consider the case when $X$ is connected. Then you can pass to a 2-fold cover $\tilde{X}\to X$ so that the foliation ${\mathcal F}$ on $X$ lifts to a transversally oriented foliation on $\tilde{X}$. See Proposition 3.5.1 of 
A. Candel, L. Conlon, "Foliations, I", Springer Verlag, 1999. 
You should read this book (and, maybe, its sequel, "Foliations, II") if you want to learn more about foliations. 
Then $\chi(\tilde{X})=0$. Thus, $\chi(X)=0$ too. Now, recall that a smooth compact connected manifold admits a nonvanishing vector field if and only if it has zero Euler characteristic. Thus, $X$ itself also admits a nonvanishing vector field. 
Incidentally, Bill Thurston proved in 1976 (Annals of Mathematics) that the converse is also true: Zero Euler characteristic for a compact connected manifold implies existence of a smooth codimension 1 foliation. This converse is much  harder. 
